PDF Outputting Duplicating Subsections

Never ignore error messages.

Your document produces

Runaway argument? { ! Paragraph ended before \@sect was complete. <to be read again> \par l.28 ...lity that at least 10 tails are flipped? } ? 

because there are blank lines (paragraph ends) inside \section. Overleaf tries to scroll past errors and show a final pdf but any such document is only for debugging, it is not intended to be usable.

\section (and \subsection etc) take the section heading as argument, not the whole section. I am not sure you really want sections here, perhaps simply a list, but I minimised the changes and kept the sectioning, but just using the single line heading in each case. Also make use of the amsmath package that you are loading to make multi-line math displays, rather than just paragraphs with unaligned inline math.

\documentclass{article} \usepackage{amsmath} \usepackage{amssymb} \newcommand{\var}{\operatorname{Var}} \newcommand{\sd}{\operatorname{SD}} \newcommand{\bin}{\operatorname{Bin}} \newcommand{\geo}{\operatorname{Geo}} \newcommand{\bern}{\operatorname{Bern}} \newcommand{\nb}{\operatorname{NB}} \newcommand{\gam}{\operatorname{Gamma}} \newcommand{\Exp}{\operatorname{Exp}} \begin{document} \begingroup \centering \LARGE STM4PSD-T5\\ \LARGE Assignment 2 - Part 1\\[0.5em] \large \today\\[0.5em] \large Ace Taylor\par \large 22812081\par \endgroup \section*{Question 1A} A fake, weighted coin has a 56 chance of flipping heads. The coin is flipped exactly 25 times. What is the expected number of tails, and what is the probability that at least 10 tails are flipped? \subsection*{Distribution needed: Binomial distribution} Formula: $X \sim \bin(25,0.44)$ - this is because there's 0.56 chance of being heads therefore the probability of tails is 1 - probability of being heads or $1 - 0.56$ which comes to $0.44$ \begin{align*} n &= 25\\ p &= 0.44\\ k&= 10 \end{align*} \section*{The expected number of tails = $E(X) = np$} $E(X) = 25 \times 0.44$ $E(X) = 11$ You can therefore expect 11 out of the 25 flips to land on tails \section{*Let a Successful attempt be when a tails is flipped} Find: $P (X \ge 10)$ OR $1 - P(X < 10)$ Probability mass function: $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$ Step 1. Calculate $P (X = K) for k = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ \[ P(X = k) = \binom{25}{k}0.44^k \times 0.56^{25 - k} \] \subsection*{For $k = 0$} \begin{align*} P(X = 0) &= \binom{25}{0} \times 0.44 ^ 0 \times (1 - 0.44) ^ {25}\\ &= \frac{25!}{25! \times 0!} \times 0.44 ^ 0 \times 0.56 ^ {25}\\ &=1 \times 1 \times 0.56^{25}\\ &= 0.000032 \end{align*} \subsection*{for $k = 1$} \begin{align*} P(X = 1) &= \binom{25}{1} \times 0.44 ^ 1 \times (1 - 0.44) ^ {25 - 1}\\ &= \frac{25!}{24! \times 1!} \times 0.44^1 \times 0.56^{24}\\ &=25 \times 0.44 \times 0.000057\\ &=0.000397 \end{align*} \subsection*{for $k = 2$} \begin{align*} P(X = 2) &= \binom{25}{2} \times 0.44 ^ 2 \times (1 - 0.44) ^ {25 - 2}\\ &=\frac{25!}{23! \times 2!} \times 0.44 ^2 \times 0.56 ^ {23}\\ &=300 \times 0.1936 \times 0.000098\\ &=0.003037 \end{align*} \subsection*{for $k = 3$} \begin{align*} P(X = 3) &= \binom{25}{3} \times 0.44 ^ 3 \times (1 - 0.44) ^ {25 - 3}\\ &=\frac{25!}{22! \times 3!} \times 0.44 ^ 3 \times 0.56 ^ {22}\\ &=2300 \times 0.085184 \times 0.00000288493\\ &=0.00056522642 \end{align*} \subsection*{for $k = 4$} \begin{align*} P(X = 4) &= \binom{25}{4} \times 0.44 ^ 4 \times (1 - 0.44) ^ {25 - 4}\\ &=\frac{25!}{21! \times 4!} \times 0.44 ^ 4 \times 0.56 ^ {21}\\ &=12650 \times 0.03748096 \times 0.00000515167\\ &=0.00244258264 \end{align*} \subsection*{for $k = 5$} \begin{align*} P(X = 5) &= \binom{25}{5} \times 0.44 ^ 5 \times (1 - 0.44) ^ {25 - 5}\\ &=\frac{25!}{20! \times 5!} \times 0.44 ^ 5 \times 0.56 ^ {20}\\ &=53130 \times 0.0164916224 \times 0.00000919942\\ &=0.00806053086 \end{align*} \subsection*{For $k = 6$} \begin{align*} P(X = 6) &= \binom{25}{6} \times 0.44 ^ 6 \times (1 - 0.44)^{25 - 6}\\ &=\frac{25!}{19! \times 6!} \times 0.44 ^ 6 \times 0.56 ^ {19}\\ &=177100 \times 0.00725631385 \times 0.00001642753\\ &=0.02111090681 \end{align*} \subsection*{for $K = 7$} \begin{align*} P(X = 7) &= \binom{25}{7} \times 0.44 ^7 \times (1 - 0.44) ^ {25 - 7}\\ &=\frac{25!}{18! \times 7!} \times 0.44 ^ 7 \times 0.56 ^ {18}\\ &=480700 \times 0.00319277809 \times 0.00002933489\\ &=0.045022263 \end{align*} \end{document}